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Find a set A with a order relation such that:

$$\forall a, b,c \in A, \inf({\sup({a,c}),b}) = \sup({\inf({a,b}),\inf({a,c})})$$

It's easy to find a set A of two or one element that satisfies this, but i'm in problem to find a set A with at least 3 elements satisfying this property (considering a ordered set A, i.e., all elements of A are real numbers, for example).

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    $\begingroup$ You're sure it's not meant to be $\inf({\sup({b,c}),a}) = \sup({\inf({a,b}),\inf({a,c})})$, which should remind you of $(B\cap C)\cup A = (B\cap A)\cup (C\cap A)$? $\endgroup$
    – JMoravitz
    May 15 at 13:33
  • $\begingroup$ no, it's in fact what I had written $\endgroup$ May 15 at 13:37
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    $\begingroup$ Are you sure you can find it with 2 elements? Notice that with 2 elements, either those infs and sups don't always exist or those elements are related, for example, $A=\{0,1\}$ with $0<1$. In this example, with $a=0$ and $b=c=1$, we obtain $1=0$, a contradiction. $\endgroup$
    – amrsa
    May 15 at 13:54
  • $\begingroup$ Indeed, if those infs and sups always exist (in which case, the ordered set is called a lattice), then the only examples you can find are the singletons. I suspect it must be as JMoravitz asked, in which case any distributive lattice will do. $\endgroup$
    – amrsa
    May 15 at 13:56
  • $\begingroup$ You are right, I didn't find with 2 elements too. But the question is supposed to be correct. $\endgroup$ May 15 at 13:58

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